Question

A student is given 2 solutions: 15.0 ml of a 0.125 M KOH solution and 12.0 ml of a 0.175 M H₂SO₄ solution. Calculate the following:

a) The amount of potassium sulfate produced.

b) What is the limiting reactant? What is the excess reactant?

c) How many grams of the excess reactants are in excess?

Answer 1

2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Moles of KOH = $\frac{15}{1000}×0.125=1.875×10^{-3}moles$

Amount of $K_2SO_4=\frac{1.875×10^{-3}}{2}=9.375×10^{-4}$

Limiting reactant is $KOH$

Excess reactant is $H_2SO_4$